Question

If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/mol D) 2.72 x 10 5 kJ/mol E) 2.72 x 10 9 kJ/mol Group of answer choices

Answer 1

Answer: The energy of the complex is $2.72\times 10^2kJ$

Explanation:

To calculate the energy of the complex, we use the equation given by Planck which is:

$\Delta E=\frac{N_Ahc}{\lambda}$

where,

$\lambda$ = Wavelength of the complex = $440nm=4.40\times 10^{-7}m$    (Conversion factor:  $1m=10^9nm$ )

h = Planck's constant = $6.624\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

$\Delta E$ = energy of the complex

Putting values in above equation, we get:

$\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ$

Conversion factor used:  1 kJ = 1000 J

Hence, the energy of the complex is $2.72\times 10^2kJ$